blocks|key|389|text|使用doc.id|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|390|users.get().then(snapshot+=>+{
++++snapshot.forEach(doc+=>+{+++++++++++
++++++++count%2B%2B;
++++++++allUserDocs[doc.id]+=+doc.data();++++++++++++
++++});

++++allUserDocs.count+=+count;
++++res.status(200).send(allUserDocs);++++++++++
})|code-block|syntax|javascript|391|entityMap^0|2|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Use <code>doc.id</code>

<pre><code>users.get().then(snapshot =&gt; {
 snapshot.forEach(doc =&gt; { 
 count++;
 allUserDocs[doc.id] = doc.data(); 
 });

 allUserDocs.count = count;
 res.status(200).send(allUserDocs); 
})
</code></pre>

blocks|key|280|text|有一个listDocuments方法可以检索有子集合的所有文档，不管是否缺失。这是解释它的page+in+the+docs。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|281|下面的代码可能就是您要查找的内容：|282|let+collectionRef+=+firestore.collection('col');

return+collectionRef.listDocuments().then(documentRefs+=>+{
+++return+firestore.getAll(...documentRefs);
}).then(documentSnapshots+=>+{
+++for+(let+documentSnapshot+of+documentSnapshots)+{
++++++if+(documentSnapshot.exists)+{
++++++++console.log(`Found+document+with+data:+${documentSnapshot.id}`);
++++++}+else+{
++++++++console.log(`Found+missing+document:+${documentSnapshot.id}`);
++++++}
+++}
});|code-block|syntax|javascript|283|​|284|您不会关心docRef是否存在。|285|然而，使用空文档听起来并不是一个好的解决方案。你追求的架构逻辑是什么，在这种架构中，用户可以是空的，但他们下面的消息仍然很重要？如果您仍然需要访问它们，您可以添加一个布尔变量来确定用户是否处于活动状态，而不是留下一个空白文档。|286|entityMap|0|LINK|mutability|MUTABLE|url|https://googleapis.dev/nodejs/firestore/latest/CollectionReference.html#listDocuments^0|19|G|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|6|7|12|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|13|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|C|$]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|C|$]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|C|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]]]

There is a listDocuments method that retrieves all documents, missing or not, that have a subcollection. Here's the <a href="https://googleapis.dev/nodejs/firestore/latest/CollectionReference.html#listDocuments" rel="nofollow noreferrer">page in the docs</a> that explains it.
Something like this might be what you are looking for:
<div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>let collectionRef = firestore.collection('col');

return collectionRef.listDocuments().then(documentRefs =&gt; {
 return firestore.getAll(...documentRefs);
}).then(documentSnapshots =&gt; {
 for (let documentSnapshot of documentSnapshots) {
 if (documentSnapshot.exists) {
 console.log(`Found document with data: ${documentSnapshot.id}`);
 } else {
 console.log(`Found missing document: ${documentSnapshot.id}`);
 }
 }
});</code></pre>
</div>
</div>

You would not care whether the docRef exists or not.
Nevertheless, it does not sound like a good solution to have empty documents. What is the logic you were pursuing with an architecture where users can be empty, but messages underneath them still matter? Maybe if you still need to access them you can add a boolean variable to determine if the user is active or not, instead of leaving a blank document.

I have collection of documents where the id of the doc is the users id.
Lets call these user documents.

Each "user document" contains a subcollection of chat messages. But not all "user documents" contains any fields (data, other than the subcollection).

I wan't to return all the doc in the collection that don't have any fields, but have a subcollection, but I seems this is not possible?

<pre><code>var allUserDocs = {}, 
 count = 0,
 users = firestore.collection("users");

users.get().then(snapshot =&gt; {
 snapshot.forEach(doc =&gt; { 
 count++;
 allUserDocs[count] = doc.data(); 
 });

 allUserDocs.count = count;
 res.status(200).send(allUserDocs); 
})
</code></pre>

this code only returns the docs that contains fields, not the docs that only have a subcollection? Is there a way to return all?

How can i get a list of all document ids in the collection? both empty and non-empty ones? or how can I add a field to all the docs without fields if i cant access them?

Firestore can't get empty docs

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个文档集合，其中文档的id是用户id。让我们称这些为用户文档。每个“用户文档”都包含聊天消息的子集。但并不是所有的“用户文档”都包含任何字段(数据，而不是子集合)。我不想返回集合中没有任何字段但有子集合的所有文档，但这似乎是不可能的。var allUserDocs = {},     count = 0,    ...

问Firestore无法获取空文档
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